Abstract
D. Marshall and S. Rohde have recently shown that there exists C0 > 0
so that the Loewner equation generates slits whenever the driving term
is H¨older continuous with exponent 1
2 and norm less than C0 [11]. In this
paper, we show that the maximal value for C0 is 4.
Results & Lemmas (10)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Theorem 1.
Theorem 1. If Gt is a quasislit-halfplane for all t, then λ ∈Lip( 1 2). Conversely, there exists C0 such that if the driving term λ ∈Lip( 1…
Theorem 1. If Gt is a quasislit-halfplane for all t, then λ ∈Lip( 1 2). Conversely, there exists C0 such that if the driving term λ ∈Lip( 1 2) with ∥λ∥1 2 < C0, then Gt is a quasislit-halfplane for all t. Although they work with the technically more challenging disc version of the Loewner equation, their techniques carry over to prove the result in the halfplane version as well. In the remainder of this paper, working with the halfplane version of Loewner’s equation, we will show that the maxima
Theorem 2.
Theorem 2. If λ ∈Lip( 1 2) with ∥λ∥1 2 < 4, then the domains Gt generated by λ are quasislit-halfplanes. Further, for each c ≥4, there…
Theorem 2. If λ ∈Lip( 1 2) with ∥λ∥1 2 < 4, then the domains Gt generated by λ are quasislit-halfplanes. Further, for each c ≥4, there exists a driving term λ ∈Lip( 1 2) with ∥λ∥1 2 = c so that λ does not generate slit-halfplanes. We will see examples of this in the next section. Similar examples were discovered independently by L. Kadanoff, W. Kager, and B. Nienhuis [8]. Their work also includes descriptions and pictures of the generated domains. There is another version of the Loewner equation
Theorem 3.
Theorem 3. If ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4, then ft(H) is a quasislit-halfplane for all t, where ft are the maps generated by ξ. When the…
Theorem 3. If ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4, then ft(H) is a quasislit-halfplane for all t, where ft are the maps generated by ξ. When the singularity catches solutions Let λ ∈Lip( 1 2) and suppose that the domains Gt generated by λ are slit- halfplanes. Then the maps gt extend continuously to R \ {λ(0)}. Thus for each x0 ∈R \ {λ(0)}, x(t) := gt(x0) is a solution to the following real-valued initial value problem: ∂ ∂tx(t) = 2 x(t) −λ(t), (4)
Lemma 1.
Lemma 1. Let λ ∈Lip( 1 2) with λ(0) = 0 and let x0 > 0. Suppose that x(t) is a solution to (4) and that x(1) = λ(1). Then ∥λ∥1 2 ≥4.
Lemma 1. Let λ ∈Lip( 1 2) with λ(0) = 0 and let x0 > 0. Suppose that x(t) is a solution to (4) and that x(1) = λ(1). Then ∥λ∥1 2 ≥4.
Lemma 2.
Lemma 2. Let λ ∈Lip( 1 2) with λ(0) = 0 and ∥λ∥1 2 < 4. Then there exists ǫ = ǫ(∥λ∥1 2 ) > 0 so that x(1) −λ(1) > ǫ, where x(t) is the…
Lemma 2. Let λ ∈Lip( 1 2) with λ(0) = 0 and ∥λ∥1 2 < 4. Then there exists ǫ = ǫ(∥λ∥1 2 ) > 0 so that x(1) −λ(1) > ǫ, where x(t) is the solution to (4) with x0 > 0.
Corollary 1.
Corollary 1. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4 and ξ(0) = 0. Suppose that x(t) is a solution to (6), with x0 ̸= 0. Then K1x2 0 ≤T (x0) ≤K2x2…
Corollary 1. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4 and ξ(0) = 0. Suppose that x(t) is a solution to (6), with x0 ̸= 0. Then K1x2 0 ≤T (x0) ≤K2x2 0, where 0 < Ki = Ki(∥ξ∥1 2 ) < ∞. 9
Lemma 3.
Lemma 3. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4. For each T > 0, there exist exactly two real numbers x0, ˆx0 so that x(T ) = ˆx(T ) = ξ(T ).
Lemma 3. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4. For each T > 0, there exist exactly two real numbers x0, ˆx0 so that x(T ) = ˆx(T ) = ξ(T ).
Lemma 4.
Lemma 4. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4 and ξ(0) = 0. There exists some constant A0 > 0, depending only on ∥ξ∥1 2, so that if 0 ≤x < y <…
Lemma 4. Let ξ ∈Lip( 1 2) with ∥ξ∥1 2 < 4 and ξ(0) = 0. There exists some constant A0 > 0, depending only on ∥ξ∥1 2 , so that if 0 ≤x < y < z with y −x = z −y, then 1 A0 ≤φ(x) −φ(y) φ(y) −φ(z) ≤A0. (7) To prove this lemma, we will need the following.
Lemma 5.
Lemma 5. Let c < 4 and 0 < ǫ < 1. Then there exists δ > 0 so that φ(β) φ(α) ≥1 + δ for non-zero α and β satisfying β α ≥1 + ǫ and for any…
Lemma 5. Let c < 4 and 0 < ǫ < 1. Then there exists δ > 0 so that φ(β) φ(α) ≥1 + δ for non-zero α and β satisfying β α ≥1 + ǫ and for any Lip( 1 2) driving term ξ with ∥ξ∥1 2 ≤c.
Lemma 6.
Lemma 6. H γ[0, T ] is a quasislit-halfplane if and only if there is a constant 1 ≤M < ∞such that 1 M ≤ x −ξ(0) ξ(0) −φ(x) ≤M for all x >…
Lemma 6. H \ γ[0, T ] is a quasislit-halfplane if and only if there is a constant 1 ≤M < ∞such that 1 M ≤ x −ξ(0) ξ(0) −φ(x) ≤M for all x > ξ(0) and 1 M ≤φ(x) −φ(y) φ(y) −φ(z) ≤M whenever ξ(0) ≤x < y < z with y −x = z −y. Furthermore, the quasislit constant K of H \ γ[0, T ] depends on M only.